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I found a solution to this online already but I'm curious to see why my method specifically doesn't work.
Let $X_1,X_2$ be independent r.v.s with parameter $\lambda$.

Then let $T = X_1 - X_2$.
We have $$\mathbb{P}(T\leq t) = \mathbb{P}(X_1 \leq t + X_1) \\ = \int_{0}^\infty \mathbb{P}(X_1\leq t + X_1 | X_1 = x) f_{X_1}(x)\ dx\\ = \int_{0}^\infty \mathbb{P}(X_1 \leq t + x | X_1 = x)f_{X_1}(x) \ dx \\ = \int_0^\infty \mathbb{P}(X_1 \leq t + x)f_{X_1}(x) \ dx \\ = \int_0^\infty (1- e^{-\lambda (t+x)})f_{X_1}(x) \ dx \\ = \int_0^\infty f_{X_1}(x) - \lambda e^{-\lambda x} e^{-\lambda t - \lambda x}\ dx \\ =1 - \int_0^\infty \lambda e^{-2\lambda x}e^{-\lambda t}\ dx \\ = 1 - e^{-\lambda t}\frac{1}{2}\int_0^\infty 2\lambda e^{-2\lambda x} \ dx \\ = 1 - \frac{1}{2}e^{-\lambda t}$$ Differentiating, I get that $$f_T(t) = \frac{\lambda}{2}e^{-\lambda t} $$ which doesn't seem like a density..

OneGapLater
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  • Where is $X_2$ ? You conditioned on $X_1$ and finding cdf of $X_1$ itself. – StubbornAtom Jun 17 '18 at 12:46
  • Think about the possible values $T$ can have, and write down the range of the cdf of $T$. Nothing will seem like a density if you do not mention the support of the random variable. – StubbornAtom Jun 17 '18 at 12:54

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